Integrand size = 30, antiderivative size = 269 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=-\frac {4 \left (d^2 e-b d f^2+a e f^2\right )}{\left (2 d e-b f^2\right )^2 \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}-\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\left (2 d e-b f^2\right )^2 \left (b f^2+2 e \left (e x+f \sqrt {a+\frac {x \left (b f^2+e^2 x\right )}{f^2}}\right )\right )}+\frac {3 f^2 \left (4 a e^2-b^2 f^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}}}{\sqrt {2 d e-b f^2}}\right )}{\sqrt {2} \sqrt {e} \left (2 d e-b f^2\right )^{5/2}} \]
3/2*f^2*(-b^2*f^2+4*a*e^2)*arctanh(2^(1/2)*e^(1/2)*(d+e*x+f*(a+b*x+e^2*x^2 /f^2)^(1/2))^(1/2)/(-b*f^2+2*d*e)^(1/2))/(-b*f^2+2*d*e)^(5/2)*2^(1/2)/e^(1 /2)-4*(a*e*f^2-b*d*f^2+d^2*e)/(-b*f^2+2*d*e)^2/(d+e*x+f*(a+b*x+e^2*x^2/f^2 )^(1/2))^(1/2)-f^2*(-b^2*f^2+4*a*e^2)*(d+e*x+f*(a+b*x+e^2*x^2/f^2)^(1/2))^ (1/2)/(-b*f^2+2*d*e)^2/(b*f^2+2*e*(e*x+f*(a+x*(b*f^2+e^2*x)/f^2)^(1/2)))
Time = 2.41 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.47 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\frac {b^2 f^4 \left (5 d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )-4 b e f^2 \left (d^2+a f^2-2 d \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )-4 e^2 \left (2 d^2 \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )+a f^2 \left (d+3 e x+3 f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}{\left (-2 d e+b f^2\right )^2 \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}} \left (b f^2+2 e \left (e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}\right )\right )}-\frac {6 \sqrt {2} a e^{3/2} f^2 \arctan \left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{\left (-2 d e+b f^2\right )^{5/2}}+\frac {3 b^2 f^4 \arctan \left (\frac {\sqrt {2} \sqrt {e} \sqrt {d+e x+f \sqrt {a+x \left (b+\frac {e^2 x}{f^2}\right )}}}{\sqrt {-2 d e+b f^2}}\right )}{\sqrt {2} \sqrt {e} \left (-2 d e+b f^2\right )^{5/2}} \]
(b^2*f^4*(5*d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) - 4*b*e*f^2*(d^2 + a*f^2 - 2*d*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)])) - 4*e^2*(2*d^2*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]) + a*f^2*(d + 3*e*x + 3*f*Sqrt[a + x*(b + (e^2*x)/f^2)])))/((-2*d*e + b*f^2)^2*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^ 2*x)/f^2)]]*(b*f^2 + 2*e*(e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]))) - (6*Sq rt[2]*a*e^(3/2)*f^2*ArcTan[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)]])/Sqrt[-2*d*e + b*f^2]])/(-2*d*e + b*f^2)^(5/2) + (3*b^2* f^4*ArcTan[(Sqrt[2]*Sqrt[e]*Sqrt[d + e*x + f*Sqrt[a + x*(b + (e^2*x)/f^2)] ])/Sqrt[-2*d*e + b*f^2]])/(Sqrt[2]*Sqrt[e]*(-2*d*e + b*f^2)^(5/2))
Time = 0.48 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2541, 1192, 1582, 27, 359, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2541 |
\(\displaystyle 2 \int \frac {e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}{\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^{3/2} \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\) |
\(\Big \downarrow \) 1192 |
\(\displaystyle 4 \int \frac {e d^2-b f^2 d+a e f^2+e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )^2-\left (2 d e-b f^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}{\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right ) \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )^2}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}\) |
\(\Big \downarrow \) 1582 |
\(\displaystyle 4 \left (\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{4 \left (2 d e-b f^2\right )^2 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}-\frac {\int -\frac {2 e^2 \left (4 \left (2 d e-b f^2\right ) \left (e d^2-b f^2 d+a e f^2\right )-\left (3 b^2 f^4-4 a e^2 f^2-8 b d e f^2+8 d^2 e^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}{\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right ) \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}}{8 e^2 \left (2 d e-b f^2\right )^2}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \left (\frac {\int \frac {4 \left (2 d e-b f^2\right ) \left (e d^2-b f^2 d+a e f^2\right )-\left (3 b^2 f^4-4 a e^2 f^2-8 b d e f^2+8 d^2 e^2\right ) \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}{\left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right ) \left (-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}}{4 \left (2 d e-b f^2\right )^2}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{4 \left (2 d e-b f^2\right )^2 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}\right )\) |
\(\Big \downarrow \) 359 |
\(\displaystyle 4 \left (\frac {3 f^2 \left (4 a e^2-b^2 f^2\right ) \int \frac {1}{-b f^2+2 d e-2 e \left (d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}\right )}d\sqrt {d+e x+f \sqrt {\frac {e^2 x^2}{f^2}+b x+a}}-\frac {4 \left (a e f^2-b d f^2+d^2 e\right )}{\sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}}{4 \left (2 d e-b f^2\right )^2}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{4 \left (2 d e-b f^2\right )^2 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}\right )\) |
\(\Big \downarrow \) 221 |
\(\displaystyle 4 \left (\frac {\frac {3 f^2 \left (4 a e^2-b^2 f^2\right ) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {e} \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{\sqrt {2 d e-b f^2}}\right )}{\sqrt {2} \sqrt {e} \sqrt {2 d e-b f^2}}-\frac {4 \left (a e f^2-b d f^2+d^2 e\right )}{\sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}}{4 \left (2 d e-b f^2\right )^2}+\frac {f^2 \left (4 a e^2-b^2 f^2\right ) \sqrt {f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x}}{4 \left (2 d e-b f^2\right )^2 \left (-2 e \left (f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}+d+e x\right )-b f^2+2 d e\right )}\right )\) |
4*((f^2*(4*a*e^2 - b^2*f^2)*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2] ])/(4*(2*d*e - b*f^2)^2*(2*d*e - b*f^2 - 2*e*(d + e*x + f*Sqrt[a + b*x + ( e^2*x^2)/f^2]))) + ((-4*(d^2*e - b*d*f^2 + a*e*f^2))/Sqrt[d + e*x + f*Sqrt [a + b*x + (e^2*x^2)/f^2]] + (3*f^2*(4*a*e^2 - b^2*f^2)*ArcTanh[(Sqrt[2]*S qrt[e]*Sqrt[d + e*x + f*Sqrt[a + b*x + (e^2*x^2)/f^2]])/Sqrt[2*d*e - b*f^2 ]])/(Sqrt[2]*Sqrt[e]*Sqrt[2*d*e - b*f^2]))/(4*(2*d*e - b*f^2)^2))
3.5.83.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_.) + (b_.)*(x_) + (c _.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Simp[2 Subst[Int[(g + h*x^n)^p*((d ^2*e - (b*d - a*e)*f^2 - (2*d*e - b*f^2)*x + e*x^2)/(-2*d*e + b*f^2 + 2*e*x )^2), x], x, d + e*x + f*Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c, d, e , f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
\[\int \frac {1}{{\left (d +e x +f \sqrt {a +b x +\frac {e^{2} x^{2}}{f^{2}}}\right )}^{\frac {3}{2}}}d x\]
Leaf count of result is larger than twice the leaf count of optimal. 696 vs. \(2 (239) = 478\).
Time = 0.79 (sec) , antiderivative size = 1456, normalized size of antiderivative = 5.41 \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\text {Too large to display} \]
[1/4*(3*(a*b^2*f^6 + 4*a*d^2*e^2*f^2 - (b^2*d^2 + 4*a^2*e^2)*f^4 + (b^3*f^ 6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*f^4)*x)*sqrt(-2*b*e*f^2 + 4*d* e^2)*log(-b^2*f^4 + 4*(b*d*e - a*e^2)*f^2 - 4*(b*e^2*f^2 - 2*d*e^3)*x - 2* (2*sqrt(-2*b*e*f^2 + 4*d*e^2)*e*f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(-2*b*e*f^2 + 4*d*e^2)*(b*f^2 + 2*e^2*x))*sqrt(e*x + f*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d) + 4*(b*e*f^3 - 2*d*e^2*f)*sqrt((b*f^2*x + e^2* x^2 + a*f^2)/f^2)) + 4*(a*b^2*e*f^6 - 8*d^4*e^3 - (5*b^2*d^2*e - 2*a*b*d*e ^2)*f^4 + 2*(7*b*d^3*e^2 - 4*a*d^2*e^3)*f^2 + 2*(b^2*e^3*f^4 - 4*b*d*e^4*f ^2 + 4*d^2*e^5)*x^2 + (b^3*e*f^6 - 4*d^3*e^4 - 2*(4*b^2*d*e^2 - 3*a*b*e^3) *f^4 + 2*(7*b*d^2*e^3 - 6*a*d*e^4)*f^2)*x + 2*(2*d^3*e^3*f + (2*b^2*d*e - 3*a*b*e^2)*f^5 - (5*b*d^2*e^2 - 6*a*d*e^3)*f^3 - (b^2*e^2*f^5 - 4*b*d*e^3* f^3 + 4*d^2*e^4*f)*x)*sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2))*sqrt(e*x + f* sqrt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d))/(a*b^3*e*f^8 + 8*d^5*e^4 - (b^ 3*d^2*e + 6*a*b^2*d*e^2)*f^6 + 6*(b^2*d^3*e^2 + 2*a*b*d^2*e^3)*f^4 - 4*(3* b*d^4*e^3 + 2*a*d^3*e^4)*f^2 + (b^4*e*f^8 - 8*b^3*d*e^2*f^6 + 24*b^2*d^2*e ^3*f^4 - 32*b*d^3*e^4*f^2 + 16*d^4*e^5)*x), -1/2*(3*(a*b^2*f^6 + 4*a*d^2*e ^2*f^2 - (b^2*d^2 + 4*a^2*e^2)*f^4 + (b^3*f^6 + 8*a*d*e^3*f^2 - 2*(b^2*d*e + 2*a*b*e^2)*f^4)*x)*sqrt(2*b*e*f^2 - 4*d*e^2)*arctan(1/2*sqrt(e*x + f*sq rt((b*f^2*x + e^2*x^2 + a*f^2)/f^2) + d)*(sqrt(2*b*e*f^2 - 4*d*e^2)*f*sqrt ((b*f^2*x + e^2*x^2 + a*f^2)/f^2) - sqrt(2*b*e*f^2 - 4*d*e^2)*(e*x + d)...
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\int \frac {1}{\left (d + e x + f \sqrt {a + b x + \frac {e^{2} x^{2}}{f^{2}}}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\int { \frac {1}{{\left (e x + \sqrt {b x + \frac {e^{2} x^{2}}{f^{2}} + a} f + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\left (d+e x+f \sqrt {a+b x+\frac {e^2 x^2}{f^2}}\right )^{3/2}} \, dx=\int \frac {1}{{\left (d+e\,x+f\,\sqrt {a+b\,x+\frac {e^2\,x^2}{f^2}}\right )}^{3/2}} \,d x \]